2.1.193 problem 195

Internal problem ID [9041]
Book : Collection of Kovacic problems
Section : section 1
Problem number : 195
Date solved : Thursday, December 12, 2024 at 10:00:03 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }-4 t y^{\prime }+\left (4 t^{2}-2\right ) y&=0 \end{align*}

Solved as second order ode using Kovacic algorithm

Time used: 0.099 (sec)

Writing the ode as

\begin{align*} y^{\prime \prime }-4 t y^{\prime }+\left (4 t^{2}-2\right ) y &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= 1 \\ B &= -4 t\tag {3} \\ C &= 4 t^{2}-2 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(t) &= y e^{\int \frac {B}{2 A} \,dt} \end{align*}

Then (2) becomes

\begin{align*} z''(t) = r z(t)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {0}{1}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 0\\ t &= 1 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= 0 \tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(t)\) then \(y\) is found using the inverse transformation

\begin{align*} y &= z \left (t \right ) e^{-\int \frac {B}{2 A} \,dt} \end{align*}

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition
3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.193: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - -\infty \\ &= \infty \end{align*}

There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Since \(r = 0\) is not a function of \(t\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is

\[ z_1(t) = 1 \]

Using the above, the solution for the original ode can now be found. The first solution to the original ode in \(y\) is found from

\begin{align*} y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {-4 t}{1} \,dt} \\ &= z_1 e^{t^{2}} \\ &= z_1 \left ({\mathrm e}^{t^{2}}\right ) \\ \end{align*}

Which simplifies to

\[ y_1 = {\mathrm e}^{t^{2}} \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \]

Substituting gives

\begin{align*} y_2 &= y_1 \int \frac { e^{\int -\frac {-4 t}{1} \,dt}}{\left (y_1\right )^2} \,dt \\ &= y_1 \int \frac { e^{2 t^{2}}}{\left (y_1\right )^2} \,dt \\ &= y_1 \left (t\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left ({\mathrm e}^{t^{2}}\right ) + c_2 \left ({\mathrm e}^{t^{2}}\left (t\right )\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )-4 t \left (\frac {d}{d t}y \left (t \right )\right )+\left (4 t^{2}-2\right ) y \left (t \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (t \right ) \\ {} & {} & y \left (t \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t^{m}\cdot y \left (t \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & t^{m}\cdot y \left (t \right )=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )}{\sum }}a_{k} t^{k +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & t^{m}\cdot y \left (t \right )=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )+m}{\sum }}a_{k -m} t^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} t \cdot \left (\frac {d}{d t}y \left (t \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & t \cdot \left (\frac {d}{d t}y \left (t \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} k \,t^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d^{2}}{d t^{2}}y \left (t \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) t^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) t^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 2 a_{2}-2 a_{0}+\left (6 a_{3}-6 a_{1}\right ) t +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )-2 a_{k} \left (2 k +1\right )+4 a_{k -2}\right ) t^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} t \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}-2 a_{0}=0, 6 a_{3}-6 a_{1}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=a_{0}, a_{3}=a_{1}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+3 k +2\right ) a_{k +2}-4 a_{k} k -2 a_{k}+4 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}+3 k +8\right ) a_{k +4}-4 a_{k +2} \left (k +2\right )-2 a_{k +2}+4 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y \left (t \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} t^{k}, a_{k +4}=\frac {2 \left (2 k a_{k +2}-2 a_{k}+5 a_{k +2}\right )}{k^{2}+7 k +12}, a_{2}=a_{0}, a_{3}=a_{1}\right ] \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Trying a Liouvillian solution using Kovacics algorithm 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
<- Kovacics algorithm successful`
 

Maple dsolve solution

Solving time : 0.007 (sec)
Leaf size : 14

dsolve(diff(diff(y(t),t),t)-4*t*diff(y(t),t)+(4*t^2-2)*y(t) = 0, 
       y(t),singsol=all)
 

\[ y = {\mathrm e}^{t^{2}} \left (c_{2} t +c_{1} \right ) \]

Mathematica DSolve solution

Solving time : 0.031 (sec)
Leaf size : 18

DSolve[{D[y[t],{t,2}]-4*t*D[y[t],t]+(4*t^2-2)*y[t]==0,{}}, 
       y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to e^{t^2} (c_2 t+c_1) \]